Integration By Substitution

= (1/4)INT dx/(1 + tan^2 &#952 :wink:

...

Reply 2

craighay1

Using the substitution x + 2 = 2tanθ integrate 1/(x^2 + 4x + 8) dx

I think you get the x + 2 by factorising the denominator into (x+2)^2 + 4

Once I substitute in 2tanθ I don't know what to do.

Sounds like you're on the right track...

you get 4tan^2(&#952 :wink:

+4 and then using tan^2x=sec^2x-1 we have 4sec^2(&#952 :wink:

-4+4

So it leaves us with (1/4) INT 1/(sec^2(&#952 :wink:

) = (1/4) INT cos^2(&#952 :wink:

I think...

Reply 3

username74116

OP

Hmm..

I've checked the answer and working back from it i think I've found how to do it. I would never have got it though.

Factorise the denominator and sub in to get 1 / (2tan&#952 :wink:

^2 + 4

This could be written as 1 / 4(tan&#952 :wink:

^2 + 4 which corresponds to the standard integral INT arctan(x) = 1 / x^2 + 1

Remebering to squareroot the fours and dividing you get 1 / 2arctan(tan&#952 :wink:

+ c

By rearranging your given substitution to get tanθ = (x+2)/2 and substituting this in you get the final answer:

1/2 arctan ((x+2)/2) + c

Reply 4

Yeah, that's one way of doing it :smile:

I didn't recognise that so I got (1/4)INT cos^2(&#952 :wink:

instead, which is (1/8)(θ+cosθsin&#952 :wink:

and I believe this is correct as well, just an equivalent form (if I've not made an algebraic slip on the way).

Reply 5

username74116

OP

Yeah but you have to remember that you would still be integrating with respect to x and not with respect to θ.

So you can't do (1/4) INT cos^2(&#952 :wink:

dx

Reply 6

Yeah

forgot that detail... though then I don't get why you were given such a horrid substitution, just u=x+2 will do the job. (or 2u=x+2)

Reply 7

steve2005

craighay1

Using the substitution x + 2 = 2tan? integrate 1/(x^2 + 4x + 8) dx

I think you get the x + 2 by factorising the denominator into (x+2)^2 + 4

Once I substitute in 2tan? I don't know what to do.

This is a bit late...and i forgot to add the c

Picture 182.png41.7KB

Reply 8

DFranklin

18

nota bene

Yeah

forgot that detail... though then I don't get why you were given such a horrid substitution, just u=x+2 will do the job. (or 2u=x+2)

u=x+2 only does the job if you already know the integral for

\int \frac{1}{x^2+a^2}\,dx

. Whereas

x+2 = 2 \tan \theta

doesn't require that knowledge. (In fact, a similar substitution is where the result for

\int \frac{1}{x^2+a^2}\,dx

comes from in the first place).

Reply 9

DFranklin

u=x+2 only does the job if you already know the integral for

\int \frac{1}{x^2+a^2}\,dx

. Whereas

x+2 = 2 \tan \theta

doesn't require that knowledge. (In fact, a similar substitution is where the result for

\int \frac{1}{x^2+a^2}\,dx

comes from in the first place).

Yeah, I know, but OP seemed to know that integral (which I believe is FP2) and substitution is C4 so seemed like an odd substitution to give when the material is used by someone knowing the integral of arctan.

I'm probably just complaining because I messed up :biggrin:

Reply 10

Kolya

Is it essential to learn the inverse trig standard forms, such as

Unparseable latex formula:

\int \frac{1}{x^2+a^2} \dx

?

Reply 11

Lusus Naturae

Is it essential to learn the inverse trig standard forms, such as

Unparseable latex formula:

\int \frac{1}{x^2+a^2} \dx

?

Uh, I dunno - I just remember them automatically :redface:

I think it makes it easier if you remember them, I mean I can integrate that mentally if I don't have a paper/pen, and it makes it quicker. However I'd not say it is necessary, so don't worry if you go around not remembering

\int\frac{1}{\sqrt{1-x^2}}=arcsin(x)+c

Reply 12

Kolya

nota bene

Uh, I dunno - I just remember them automatically

I think it makes it easier if you remember them, I mean I can integrate that mentally if I don't have a paper/pen, and it makes it quicker. However I'd not say it is necessary, so don't worry if you go around not remembering

\int\frac{1}{\sqrt{1-x^2}}=arcsin(x)+c

I just use substitution to solve questions involving an inverse trig function, but seem to be in a minority here. :redface:

Reply 13

username21537

2

Lusus Naturae

Is it essential to learn the inverse trig standard forms, such as

?

I would suggest so if you want to take maths on past A level.

Which is why my revision period is gonna have a lot of focus on learning them

Reply 14

Lusus Naturae

I just use substitution to solve questions involving an inverse trig function, but seem to be in a minority here. :redface:

What do you do when you differentiate e.g. arcsin(x) then? First principles? That must mean a lot of work...

Sure the substitutions are fine :smile:

Just that under certain conditions we have time limits (like my test this Monday which I totally messed up due to waaaay too little time). But whatever floats your boat! I'm sure you're used to do it your way and do so quickly.

Reply 15

Kolya

nota bene

What do you do when you differentiate e.g. arcsin(x) then? First principles? That must mean a lot of work...

I should have said that I know the v. basic standard forms, and use these in differentiation - ie. with x's and 1's - so differentiating arcsin(x) would not be a problem.

OK, picking a random example to show my unusual (and poor?) methods:

\displaystyle \int \frac{4x-1}{\sqrt{x^2+16}} \ dx = \int \frac{4x}{\sqrt{x^2+16}} \ dx - \int \frac{1}{\sqrt{x^2+16}} \ dx

Here I would look at

\int \frac{4x-1}{\sqrt{x^2+16}} \ dx

, and guess the answer is

\sqrt{x^2+16}

, I would then differentiate this in my head, and mentally adjust to give the answer to

this integral as

Unparseable latex formula:

\sqrt{x^2+16

Now, relevant to the discussion, looking at

\displaystyle \int \frac{1}{\sqrt{x^2+16}} \ dx = \frac{1}{4} \int \frac{1}{\sqrt{\frac{x}{4}^2+1}} \ dx

Let shy = x/4

Unparseable latex formula:

\dispalystyle \int \frac{4chy}{\sqrt{sh^2+1}} \ dy

Everything always cancels, so in this case it gives:

\int 1 \ dy = y = arsh(\frac{x}{4})

Therefore answer is:

4\sqrt{x^2-16} - arsh(\frac{x}{4})

I question like this would take around 2-3 minutes.

Nota, should I scrap it, and try and learn the normal way? :redface:

Reply 16

DFranklin

18

Lusus Naturae

Should I scrap it, and try and learn the normal way? :redface:

The only clear scenario I can see for learning off by heart is if you have multiple choice exams; if you have only 2 minutes per question then you really might want to save the time it takes to derive the standard result.

Otherwise, as you say, it's only a few minutes, it makes sure you know what's going on as opposed to using a rote formula, and it's good practice for when you get similar but not completely standard questions requiring substitutions.

Reply 17